# Fluid mechanics

### 1 個解答

• 最愛解答

Consider point A and the nozzle, apply Bernoulli's Equation,

Pa + (1/2)p(Va)^2 = Pn + (1/2)p(Vn)^2 + pg(1.1)

where Pa and Pn are the pressures at A and at the nozzle respectively,

p os the density of water (= 1000 kg/m^3)

Va and Vn are the speeds of water at A and at the nozzle respectively

g is the acceleration due to gravity

Because the water is at free flow at the nozzle, Pn = 0

thus, (55 x 10^3) + (1000/2)(Va)^2 = (1000/2)(Vn)^2 + 1000 x 9.81 x 1.1

simplifying, 88.42 + (Va)^2 = (Vn)^2 ----------- (1)

By equation of continuity,

(Aa).(Va) = (An).(Vn)

where Aa and An are the cross-sectional areas of the pipe at point A and at the nozzle respectively.

hence, [pi(0.2/2)^2].(Va) = [pi.(0.1/2)^2].(Vn)

0.04(Va) = 0.01(Vn)

i.e. Va = Vn/4 ----------- (2)

Substitute Va from (2) into (1),

88.42 + (Vn/4)^2 = (Vn)^2

solve for (Vn)^2 gives (Vn)^2 = 94.31 (m/s)^2

By conservation of mechanical energy,

gain in potential energy of water at top of water jet = loss of kinetic energy of water at nozzle,

thus, pgh = (1/2)p(Vn)^2

h = (Vn)^2/(2g) = 94.31/(2x9.81) m = 4.81 m

• 登入以回覆解答